#869 - Reordered Power of 2

Medium • View on LeetCode • View on GitHub

Hash Table Math Sorting Counting Enumeration

Problem Description

You are given an integer n. We reorder the digits in any order (including the original order) such that the leading digit is not zero.

Return true if and only if we can do this so that the resulting number is a power of two.

Solution

/**
 * @param {number} n
 * @return {boolean}
 */
var reorderedPowerOf2 = function(n) {
  const numSignature = getDigitCount(n);
  const maxPower = Math.floor(Math.log2(10 ** 9));

  for (let i = 0; i <= maxPower; i++) {
    if (getDigitCount(1 << i) === numSignature) return true;
  }

  return false;

  function getDigitCount(num) {
    const count = Array(10).fill(0);
    while (num > 0) {
      count[num % 10]++;
      num = Math.floor(num / 10);
    }
    return count.join('');
  }
};